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3 tháng 3 2017

a) x 2  – x – 12                              b) x 3  – 64.

c) m 3 n 3   –   m 2 n   +   5 mn 2   –   5          d) 16 x 4  – 1.

6 tháng 8 2018

\(b,5x\left(x-1\right)-3x\left(1-x\right)=\left(5x+3x\right)\left(x-1\right)\)

\(c,-16a^4.b^6-24a^5.b^5-9a^6.b^4\)

\(=-a^4.b^4[\left(4b\right)^2+2.4.a.3.b+\left(3a\right)^2]\)

\(=-a^4.b^4\left(4b+3a\right)^2\)

15 tháng 7 2016

a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

c)4x+4y-x^2(x+y)

= 4(x + y) - x^2(x+y)

= (4-x^2) (x+y)

= (2-x)(2+x)(x+y)

d) y^2+y-x^2+x

= (y^2 - x^2) + (x+y)

= (y-x)(y+x)+ (x+y)

= (y-x+1) (x+y)

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

= (2x-y)(2x+y) - (2x+y)

= (2x -y -1)(2x+y)

f)9x^2-25y^2-6x+10y

31 tháng 8 2021

ko biết làm

 

Câu 5:B

Câu 4: C

Câu 3: D

Câu 2: A

Câu 1: A

19 tháng 8 2020

a/ \(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)^2\left(x^2-x+1\right)\)

b/ \(x^4-x^3-x^2+1=\left(x^4-x^3\right)-\left(x^2-1\right)=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^3-x-1\right)\)

c/ \(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1-x+1\right)\left(2x+1+x-1\right)=3x\left(x+2\right)\)

d/ \(x^4+4x^2-5=x^4-x^2+5x^2-5=x^2\left(x^2-1\right)+5\left(x^2-1\right)=\left(x^2-1\right)\left(x^2+5\right)=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

19 tháng 8 2020

a, x4 + x3 + x + 1

= x3(x + 1) + (x + 1)

= (x + 1)(x3 + 1)

= (x + 1)2(x2 - x + 1)
b, x4 - x3 - x2 + 1

= x3(x - 1) - (x - 1)(x + 1)

= (x - 1)(x3 - x - 1)
c, ( 2x + 1 )2 - ( x - 1 )2

= (2x + 1 + x - 1)(2x + 1 - x + 1)

= 3x(x + 2)
d, x4 + 4x2 - 5

= x4 - x2 + 5x2 - 5

= x2(x2 - 1) + 5(x2 - 1)

= (x2 - 1)(x2 + 5)

= (x - 1)(x + 1)(x2 + 5)

a) Ta có: \(x^2+4x+3\)

\(=x^2+x+3x+3\)

\(=x\left(x+1\right)+3\left(x+1\right)\)

\(=\left(x+1\right)\left(x+3\right)\)

b) Ta có: \(16x-5x^2-3\)

\(=-5x^2+16x-3\)

\(=-5x^2+15x+x-3\)

\(=-5x\left(x-3\right)+\left(x-3\right)\)

\(=\left(x-3\right)\left(-5x+1\right)\)

c) Ta có: \(2x^2+7x+5\)

\(=2x^2+2x+5x+5\)

\(=2x\left(x+1\right)+5\left(x+1\right)\)

\(=\left(x+1\right)\left(2x+5\right)\)

d) Ta có: \(2x^2+3x-5\)

\(=2x^2+5x-2x-5\)

\(=x\left(2x+5\right)-\left(2x+5\right)\)

\(=\left(2x+5\right)\left(x-1\right)\)

e) Ta có: \(x^3-3x^2+1-3x\)

\(=\left(x+1\right)\cdot\left(x^2-x+1\right)-3x\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)

\(=\left(x+1\right)\left(x^2-4x+1\right)\)

f) Ta có: \(x^2-4x-5\)

\(=x^2-4x+4-9\)

\(=\left(x-2\right)^2-3^2\)

\(=\left(x-2-3\right)\left(x-2+3\right)\)

\(=\left(x-5\right)\left(x+1\right)\)

g) Ta có: \(\left(a^2+1\right)^2-4a^2\)

\(=\left(a^2+1\right)^2-\left(2a\right)^2\)

\(=\left(a^2+1-2a\right)\left(a^2+1+2a\right)\)

\(=\left(a-1\right)^2\cdot\left(a+1\right)^2\)

h) Ta có: \(x^3-3x^2-4x+12\)

\(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-4\right)\)

\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

i) Ta có: \(x^4+x^3+x+1\)

\(=x^3\left(x+1\right)+\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3+1\right)\)

\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)

k) Ta có: \(x^4-x^3-x^2+1\)

\(=x^3\left(x-1\right)-\left(x^2-1\right)\)

\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)\left(x^3-x-1\right)\)

l) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)

\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)

\(=3x\left(x+2\right)\)

m) Ta có: \(x^4+4x^2-5\)

\(=x^4-x^2+5x^2-5\)

\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^2+5\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

22 tháng 9 2021

Bài 5:

a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)

\(minA=5\Leftrightarrow x=2\)

b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

Bài 4:

a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(maxM=7\Leftrightarrow x=2\)

b) \(N=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)

\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)